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Full text of "Treatise On Analysis Vol-Ii"

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(15.4.13)    Let B be a star algebra with unit element e i=- 0, and A a closed
self-adjoint subalgebra ofB containing e. Then SpA(x) = SpB(x)for all x e A.

If x is self-adjoint, this follows because SpA(x) c R and therefore all the
points of SpA(X) are frontier points in C, and we need only apply (15.2.8). On
the other hand, if x e A is invertible in B, then so is xx* (15.4.3), which is self-
adjoint and belongs to A. From above, xx* is invertible in A, hence x has
a right inverse in A. The same argument, applied to x*x9 shows that x is
invertible in A. If we replace x by x — Łe in this argument, we obtain the
relation SpA(#) = SpB(X).

(15.4.14)        (Gelfand-Neumark theorem)   Let A be a separable commuta-
tive star algebra with unit element e ^ 0. Then the Gelfand transformation
(15.3) is an isometric isomorphism of A onto <^C(X(A)) (that is to say,
p(x) = || 9x\\ = \\x\\) such that <$x* = Wxfor all xeA.

If x is a character of A, then %(x) is real for all self-adjoint x e A, by
virtue of (15.4.12) and (15.3.1 ). For each element x = xl + ix2 of A, where x1
and x2 are self-adjoint (15.4.11), we have therefore

and therefore <&x* = <$x. Next, the elements &x e <^C(X(A)) separate the
points of X(A) (7.3), for if Xi ^ fa are two distinct characters, then by defini-
tion there exists x e A such that %i(x) ^ XaW* in other words such that
(^x)(xi) ^ (<^x)(x2)- Also the constants belong to ^(A), because &e is the
constant 1. Hence by (7.3.2) the subalgebra #(A) is dense in ^C(X(A)). To
complete the proof it is enough to show that ^ is isometric, for then ^(A),
being isomorphic to A, will be a complete subspace of ^C(X(A)), and therefore
will be equal to VC(X(A)) (3.14.4).

We shall first prove the following lemma:

( In a (commutative or non commutative) star algebra A, we have
p(y) = \\y\\ for every self-adjoint y e A.

We have ||^||2 = ||/|| by (15.4.5) and therefore ||,y||2" = ||/1 by induction
on n. Hence \\y\\ = ||>>2"||2~n; letting «->>oo, the right-hand side tends to
p(y) (15.2.7), which proves the result.

Now return to the proof of (15.4.14). Let x be any element of A; then
||*||2=: \\jfx\\ = ||#(**x)|| by the lemma and (15.3.4(i)). But $(x*x) =
^x* • %x = \&x\2, and ||^*x)|| = ||*jc||2.                                       Q.E.D.A be a separable Banach algebra with unit element. Show that if a ^= A is a left